POTW Returns

If that last post did not satisfy your need for brain food, then let me mention that as of today the Problem of the Week returns. This semester’s theme: Fun With Arithmetic! What’s that? You don’t like arithmetic? Well, let’s see if you’re still saying that at the end of the term.

In general I try to choose problems that are accessible even to people in lower level classes. I go for things with a brainteaser quality to them, as opposed to problems that require knowledge of calculus or something higher. Also, I’m perfectly aware that solutions to just about any brainteaser are readily available online. But what can you do? If a problem is charming and likely to be new to most students, then that is what I care about. The students do not receive any course credit for participating, meaning that the one’s who do are precisely the one’s who tend to enjoy solving things on their own. And, frankly, if they care enough to go looking online, and then to express the answer in their own way, then I’ll take that.

So, be sure to check in every Monday for a new problem.


  1. #1 GAZZA
    September 2, 2014

    I believe I have a solution. I suspect that some numbers can be formed more than one way though.

    Does this commenting engine allow spoiler tags? (I don’t want to spoil anyone else, but assuming a few of us solve this I’d like to compare answers).

  2. #2 MNb
    September 2, 2014

    This is a fun problem indeed and not too hard. Am I right that 13 is the smallest number that doesn’t have a solution (ie takes at least 5 fours?)

  3. #3 GAZZA
    September 2, 2014

    Well I only did to 10, but now that you mention it, I have a 13 solution with only 4 fours, as well as 14-18. 19 I’m struggling with.

  4. #4 GAZZA
    September 2, 2014

    Scratch that, I have 19-22. Thinking about 23 now.

  5. #5 GAZZA
    September 2, 2014

    OK, I have everything up to 31. And I feel very strongly that 31 should be possible; I just haven’t gotten it yet.

  6. #6 eric
    September 2, 2014

    Must you use exactly four 4’s to produce each number, or can you use up to four?

  7. #7 eric
    September 2, 2014

    Gazza – I can get to both 30 and 32 using three 4’s. If you have a way of subtracting or adding 1 using only one 4, we can share info.

  8. #8 Jason Rosenhouse
    September 2, 2014

    The problem statement calls for using exactly four four’s. This actually makes some of the numbers a little bit harder to obtain. For example, it’s easy to do 8=4+4, but you have to work a little harder to come up with 8=(4+4)(4/4).

    In general there are going to be multiple ways of obtaining most of the numbers, especially for small numbers, so you shouldn’t think the solution has to be unique.

    With a bit of Googling, you can find solutions for numbers all the way up to 116, with the sole exception of 113. I won’t include the link though, since people seem to be enjoying working it out for themselves.

    Next week’s problem has a similar flavor. It is based on the mathematical card game Krypto.

  9. #9 GAZZA
    September 3, 2014

    eric – Yeah, I can get 30 and 32 with 3 4s as well, for example 4! + 4!/4 = 30, and 4 * 4 * sqrt(4) = 32. I’ve found myself stymied there though; I still can’t think of a way to get 31.

    I would however be happy to share info if you like; just don’t want to do it “unspoiled”.

  10. #10 eric
    September 3, 2014

    Gazza – I now suspect that part of the point is to have the student learn that there are many many more arithmatic functions than the ones they are familiar with.

    I looked some up, and both the sigma and aliquot sum functions will help produce the needed odd numbers. Sigma(n) is the sum of all divisors of n, and the aliquot function s(n) is the sum of all divisors of n excluding n. So…

    sigma(4) = 1+2+4 = 7
    s(4) = 1+2 = 3
    Apply as needed. 🙂

  11. #11 eric
    September 3, 2014

    Just a small add, s(squrt(4)) = 1.

  12. #12 GAZZA
    September 3, 2014

    Interesting. That certainly solves the problem, of course, but I wonder if it’s legal. The problem statement says that you can use “… whatever standard mathematical notation you like (such as fraction bars, square roots, factorials, decimal points, parentheses, and exponents) …”. The lack of an “etc” there might imply that list is intended to be exhaustive, but perhaps not.

  13. #13 Sean T
    September 4, 2014

    I can help with 32. 4^4 = 64. 4^4/4 = 16. sqrt(4) =2 . Thus,

    4^4/4 * sqrt(4) = 32.

  14. #14 Sean T
    September 4, 2014

    It’s also obvious that uniqueness of solutions is not implied. For instance, Jason gave 4/4 * (4+4) as the solution for 8. It’s pretty easy to see that 4+4+4-4 is also a solution for 8.

  15. #15 Sean T
    September 4, 2014

    For 31, one solution:


  16. #16 eric
    September 4, 2014

    @15 – maybe you need to parse the parentheses again. If that’s 4^(4-2) / 2, it equals 8. If thats [(4^4)-2] / 2, it equals 127.
    I think you were trying for [(4^3)-2]/2, but that formula requires one 4 for the 4, one 4 for each of the 2’s, and you can’t get that exponent 3 from a single 4 without using an unusual function (like one of the ones I described above).

  17. #17 eric
    September 4, 2014

    If rounding functions are allowed round(ln(4)) and round(Log(4)) both = 1, another way to get an odd digit using a single 4. And round(e^4) = 55.

    There is no ‘standard mathematical notation’ for those functions, however, so its probably cheating.

  18. #18 Sean T
    September 4, 2014


    Sorry, I had a blonde moment. I now realize that I erroneously calculated 4^4 as 64 instead of the correct 128. I will keep working on “31”.

  19. #19 Sean T
    September 4, 2014

    Replacing my incorrect “32” solution above:

    4! + 4 + sqrt(4) + sqrt(4) = 32

  20. #20 Sean T
    September 4, 2014

    “31” was bugging me so I googled it. It is indeed possible with only the functions that Jason lists in the original problem; no rounding or sigma functions are needed. I won’t post the solution unless you want me to, though, so you can continue to work on it if you want.

  21. #21 eric
    September 4, 2014

    No, do not (at least not yet), I’m still cogitating on it in my spare moments. If you want to publish it on Monday, fine by me.

  22. #22 Phil B
    September 4, 2014

    Why is 31 causing so much grief? It really is one of the easiest. Try 4 squared + 4 squared – (4/4).

  23. #23 Gary Sturgess
    September 4, 2014

    Phil – there is no “squared” function. You’d have to do something like:
    4 * 4 + 4 * 4 – 4/4
    4^sqrt(4) + 4^sqrt(4) – 4/4

    (neither of which are very efficient but are most similar to your proposal). Both of these use too many fours.

    And Sean, I’m still thinking about it too. I have a feeling once I get 31 I’ll easily get at least another 5 or 6 (32 is easy, and whatever gives 31 may well give the key to 33 as well).

  24. #24 Sean T
    September 5, 2014


    Don’t know about 33, but if you were having trouble with 29, then the answer to 31 would make 29 obvious (Hopefully I didn’t give too much away to those who are still trying to work it out).

  25. #25 Sean T
    September 5, 2014


    I just thought of a second solution (other than the one I googled) for 31. This solution would indeed yield a solution to 33 in a very obvious way. Unless someone objects, I will post both solutions on Monday.

  26. #26 Gary Sturgess
    September 5, 2014

    I have 29, but not in a way that helps me get 31. My solution for 29 is 4! + 4 + 4/4.

    As I say, up to 30 is relatively easy. And 32 is trivial, but since I don’t have 31 yet I haven’t looked beyond that.

  27. #27 eric
    September 5, 2014

    Figured out *a* way of getting 31 this morning, but it requires I concatenate the arithmetical results of operations on the four, not just concatenate fours. I’m not sure that’s allowed. Anyway, here it is:
    ((4/4) concatinate 4!)/4
    = (1 concatinate 24)/4
    = 124/4
    = 31

  28. #28 Gary Sturgess
    September 5, 2014

    Ingenious, but I’m not sure we can reasonably argue that “concatenate” is a standard arithmetic function.

  29. #29 Sean T
    September 5, 2014

    I won’t comment on the definition of “standard arithmetic functions.” However, I do know that two solutions to 31 are possible using only the operations of addition, multiplication, subtraction and division, along with square roots, exponentiation and factorials. Parentheses are also used in these two solutions.

  30. #30 eric
    September 5, 2014


    I’m not sure we can reasonably argue that “concatenate” is a standard arithmetic function.

    Oh it isn’t. But Jason’s instructions say you can do it. What they don’t do is say whether you can only do it to the original 4’s (i.e., to make 44, 444, or 4444), or whether you can also do it to other digits (i.e. sqrt(4) and sqrt(4) to make 22).

  31. #31 eric
    September 5, 2014

    @29 – grrrr…now I’m going to keep working on it…

  32. #32 Jr
    September 8, 2014

    Wikipedia actually lists a solution that works for every positive integer, as long as you accept the use of the ln-function: http://en.wikipedia.org/wiki/Four_fours#Rules

  33. #33 Sean T
    September 8, 2014


    Please let me know when you want me to post solutions for 31. Just to give a hint in case you are still struggling: here’s an easily proved algebraic formula

    a/b + 1 = (a+b)/b

  34. #34 eric
    September 8, 2014

    I’m okay with posting the monday after. Presumably Jason’s students have all put their answers in by now, and after all, it’s problem of the week, not problem of the month. 🙂

  35. #35 Sean T
    September 8, 2014

    I agree eric. First the “google” solution:

    Realizing that 6!/4! = 6*5 = 30, that gives

    6!/4! + 1 = 31. 6 can be expressed as 4 + sqrt(4), That gives
    (4+sqrt(4))!/4! + 1 = 31. Final step is how to express that 1 using only the 1 remaining 4. The answer is the formula I gave for the hint above. ((4 +sqrt(4))! + 4!)/4!. (It’s tough to render formulas on these comments. In case that’s unclear, the numerator is 6! + 4!. The denominator is 4!, with the “6” being expressed as 4+sqrt4 )

    Now for the one I came up with. I was inspired by my own “blonde moment” upthread. For those who don’t want to refer back, I somehow incorrectly calculated 4^4 as 64 instead of 128. I realized, though, that I would have a valid solution if I could express 64 correctly using two 4’s. The solution that came to me was that if I could take the 8th root of 4^(4!) = 4^24 that would give me my 64. Fortunately, the operation of nested square roots will do this. In this case, three nested radicals will get the value I want. Thus,
    sqrt(sqrt(sqrt(4^4!) = 64. If we take this value for 64 (abbreviated as just “64” in the following) it’s almost trivial to get 31: (64 – sqrt(4))/sqrt(4)

  36. #36 Sean T
    September 8, 2014

    Also note, if you did not already have solutions for 29 and 33, you could get them trivially from these two solutions for 31. Just change the addition or subtraction signs in the numerator to the opposite to generate these.

  37. #37 eric
    September 8, 2014

    Fortunately, the operation of nested square roots will do this. In this case, three nested radicals will get the value I want. Thus, sqrt(sqrt(sqrt(4^4!) = 64.

    Hmm, it seems vaguely cheating to say nth roots require no extra fours for any n…but my earlier suggestions weren’t any better 🙂

  38. #38 Sean T
    September 8, 2014


    I don’t think it’s cheating. This representation does not work for any root, only for roots that are powers of two. The square root of the square root of a quantity is the fourth root. The square root of the fourth root would be the eighth root. The square root of the eighth root would be the sixteenth root, etc. It is using standard mathematical notation only, namely the radical sign. No other digits are required to convey the intended values, so I am not sure why it would be contrary to the spirit of the problem to use nested square roots.

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